Question

The only force acting on a $ 2.0 \,kg $ body as it moves along a positive $ x $ -axis has an $ x $ -component $ F_x = - 6x $ with $ x $ in metres. The velocity at $ x = 3.0\,m $ is $ 8.0 \,m/s $ . The velocity of the body at $ x = 4.0 \,m $ is

• A. $ 6.6 \,m/s $
• B. $ 46.6 \,m/s $
• C. $ 60 \,m/s $
• D. $ 96.6 \,m/s $

Answer 1

Answer: (A)

As the body moves along the $x$-axis from
$x_i = 3.0 \,m$ to $x_f = 4.0 \,m$, the work done by the force is
$W = \int\limits_{x_i}^{x_{f}} f_{x} dx = \int\limits_{x_i}^{x_{f}} -6x dx = - 3\left(x_{f}^{2} - x_{i}^{2}\right) $
$ = -3 \left[\left(4.0\right)^{2} -\left(3.0\right)^{2}\right] 11 = -21\, J$
According to the work-kinetic energy theorem, this gives the change in the kinetic energy.
$W = \Delta K = \frac{1}{2} m(v_f^2 - v_i^2)$
where, $v_i$ is the initial velocity (at $x_i$ and $v_f$ is the final velocity (at $x_f$). The theorem yields.
$v_{f} = \sqrt{\frac{2W}{m} +v_{i}^{2}}$
$ = \sqrt{\frac{2\left(-21\right)}{2.0} +\left(8.0\right)^{2}}$
$ = 6.6\, m/s$